Answer:
decrease
Explanation:
weight
As water accumulates in the car, its speed decrease according to Newton's second law of motion
What is newton's second law of motion?Newton's second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
according to newton's second law
force = mass * acceleration
as water accumulate in the car , mass will increase
since mass and acceleration are inversely proportional to each other
hence , acceleration will decrease and speed will decrease
learn more about newton's second law:
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A machine carries a 100kg cargo to a boat at a rate of 10m/s2. The distance between the ground to the boat is 50ft. If the machine must transfer the cargo to the boat in 5 minutes, how much power should the machine exert?
Answer:
50.8 watt
Explanation:
we know that P=W÷t
W=F.S S-->distance=50 ft= 15.24 m
F=ma
=100×10=1000 N
SO W= 1000×15.24
=15240 J
NOW
P=W÷t t=5 mints = 5×60=300 sec
P=15240÷300
P=50.8 watt
What is the force applied by the ground called? A. applied force B. friction C. gravity
Answer:
gravity i think hope this helps
Explanation:
1.
Which of the follow
following
cannot be zero?
A. Distance B. Displacement
" Velocity D. Acceleration
C.
A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −0.0200 m in the same direction as the electric field. The electron's speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.04 ✕ 10−17 J. The electron continues to x = −0.210 m within the constant electric field. If there's a change in potential energy of −9.60 ✕ 10−17 J as it goes from x = 0.190 m to x = −0.210 m, find the electron's speed (in m/s) at x = −0.210 m.
Answer:
The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]
Explanation:
Given that,
Electric field [tex]E=1.50\times10^{3}\ N/C[/tex]
Distance = -0.0200
The electron's speed has fallen by half when it reaches x = 0.190 m.
Potential energy [tex]P.E=5.04\times10^{-17}\ J[/tex]
Change in potential energy [tex]\Delta P.E=-9.60\times10^{-17}\ J[/tex] as it goes x = 0.190 m to x = -0.210 m
We need to calculate the work done by the electric field
Using formula of work done
[tex]W=-eE\Delta x[/tex]
Put the value into the formula
[tex]W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))[/tex]
[tex]W=-5.04\times10^{-17}\ J[/tex]
We need to calculate the initial velocity
Using change in kinetic energy,
[tex]\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2[/tex]
[tex]\Delta K.E=\dfrac{-3mv^2}{8}[/tex]
Now, using work energy theorem
[tex]\Delta K.E=W[/tex]
[tex]\Delta K.E=\Delta U[/tex]
So, [tex]\Delta U=W[/tex]
Put the value in the equation
[tex]\dfrac{-3mv^2}{8}=-5.04\times10^{-17}[/tex]
[tex]v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}[/tex]
Put the value of m
[tex]v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}[/tex]
[tex]v=1.21\times10^{7}\ m/s[/tex]
We need to calculate the change in potential energy
Using given potential energy
[tex]\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})[/tex]
[tex]\Delta U=-4.56\times10^{-17}\ J[/tex]
We need to calculate the speed of electron
Using change in energy
[tex]\Delta U=-W=-\Delta K.E[/tex]
[tex]\Delta K.E=\Delta U[/tex]
[tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}[/tex]
Put the value into the formula
[tex]v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}[/tex]
[tex]v_{f}=1.5\times10^{7}\ m/s[/tex]
Hence, The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]
A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340 kPa. The bottle is turned upsidedown and the cap is released so that the water is rapidly forced out of the bottle. If the air in the bottle undergoes an adiabatic pressure change, what is the pressure in the bottle when the bottle is five-sixths full of air
Answer:
The value is [tex] P_G = 2.925 *10^{5} \ Pa[/tex]
Explanation:
From the question we are told that
The volume of the bottle is [tex]v = 2 \ L = 2 * 10^{-3} \ m^3[/tex]
The gauge pressure of the air is [tex]P_g = 340 \ kPa = 340 *10 ^{3} \ Pa[/tex]
Generally the volume of air before the bottle is turned upside down is
[tex]V_a = \frac{3}{4} * V[/tex]
[tex]V_a = \frac{3}{4} * 2 *10^{-3}[/tex]
[tex]V_a = 0.0015 \ m^3 }[/tex]
Generally the volume air when the bottle is turned upside-down is
[tex]V_u = \frac{5}{6} * 2 *10^{-3}[/tex]
[tex]V_u = 0.00167 \ m^3 [/tex]
From the the mathematical relation of adiabatic process we have that
[tex]P_g * V_a^r = P_G * V_u^r[/tex]
Here r is a constant with a value r = 1.4
So
[tex] 340 *10 ^{3} * 0.0015^{1.4} = P_G * 0.00167^{1.4}[/tex]
[tex] P_G = 2.925 *10^{5} \ Pa[/tex]
An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of the transverse and longitudinal waves in rock are 8.8 km/s and 5.9 km/s respectively. A seismograph records the arrival of the transverse waves 69 s before that of the longitudinal waves. Assuming the waves travel in straight lines, how far away is the center of the earthquake
Answer:
1239.216 km
Explanation:
The speed of the transverse = 8.8km/s
The speed of the longitudinal = 5.9km/s
distance = speed x time,
8.8km/s x trans_time = 5.9km/s x long_time
8.8 / 5.9 = long_time / trans_time
1.49 = long_time / trans_time
long_time = 1.49 trans_time
the transverse wave was 69s faster than longitudinal,
trans_time - long_time = 69s
trans_time - 1.49trans_time = 69s
0.49 trans_time = 69
trans_time = 69 / 0.49 = 140.82s
long_time = 140.82 - 69 = 71.82s
the distance of the earthquake;
distance = 8.8 x 140.82 = 1239.216 km
A group of engineers is preparing a satellite to land by moving it 10% closer to
Earth in each rotation. Which statement is correct about the rotational inertia of
the satellite? (1 point)
O Rotational inertia does not change because it is conserved.
Rotational inertia increases proportional to the decrease in the
radius of rotation.
Rotational inertia first decreases and then increases as the
satellite is ready to land.
O
Rotational inertia decreases proportional to the decrease in the
radius of rotation.
Answer:
Rotational inertia decreases proportional to the decrease in the radius of rotation.
Explanation:
The rotational inertia decreases proportional to the decrease in the
radius of rotation.
Rotational inertia is directly proportional to the angular momentum of the object and inversely proportional to its angular velocity.
[tex]I = \frac{L}{\omega}[/tex]
where;
I is the rotational inertiaL is the angular momentumω is the angular speedThe angular momentum is given as;
[tex]L = mvr[/tex]
where;
m is the mass of the objectv is the velocity of the objectr is the radius of the objectThe new rotational inertia equation becomes;
[tex]I = \frac{mvr}{\omega}[/tex]
From this equation, we can observe that the rotational inertia is directly proportional to the radius of the object.A 10% closer to the Earth, means a decrease in the radius of the satellite by 10%.
Thus, a decrease in the rotational inertia as well.
Learn more here: https://brainly.com/question/17285721
what would the answer be ?
An ionized oxygen molecule (O2+) at point A has charge +e and moves at 1.24 ✕ 103 m/s in the positive x-direction. A constant electric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.766 mm past A on the x-axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of an oxygen molecule is 5.31 ✕ 10−26 kg and the fundamental charge is e = 1.60 ✕ 10−19 C.)
Answer:
[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]
[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]
Explanation:
Given that:
The charge on the ionized oxygen molecule = +e
The speed of the ionized oxygen molecule with this charge is 1.24 × 10³ m/s
distance travelled by the particle before rest is d = 0.766 m
According to the third equation of motion.
[tex]v^2 = u^2 +2as[/tex]
[tex]v^2 = u^2 +2(\dfrac{-eE}{m}) s[/tex]
[tex]0^2= u^2 +2(\dfrac{-eE}{m}) s[/tex]
[tex]E = \dfrac{mu^2}{2e* \ s}[/tex]
[tex]E = \dfrac{5.31 *10^{-26}* (1.24*10^3)^2}{2*1.6*10^{-19}*0.766*10^{-3}}[/tex]
[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]
Thus, the electric field shows to be in the negative x-direction.
The potential difference between point A and B now is:
[tex]\Delta V = E.d \\ \\ V_A - V_B = 3.33 \times 10^2 \times 0.766 \times 10^{-3}[/tex]
[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]
A woman on a snowmobile moving with a constant velocity east down the road fires a flare straight upward and the snowmobile continues to move with a constant
velocity as the flare is in the air. Assume no air resistance. Where will the flare land?
A) it will hit the person on the snowmobile who fired the flare
B) Somewhere behind the snowmobile, depends on velocity
C) in front of the snowmobile
D) Impossible to know
E) behind the snowmobile in exactly the same location (relative to the ground) from which it was fired
The volume of water and an egg in a graduated beaker is 200mL. After the egg is removed the volume of the water is found to be 125mL . What is the volume of the egg in cm ^3
Answer:
75 cm³
Explanation:
The volume of the egg is equal to the volume of the egg and water minus the volume of the water.
V = 200 mL − 125 mL
V = 75 mL
V = 75 cm³
Answer:
75 cm
Explanation:
The equation would be 200ml -125 ml thus finding the volume of the egg.
Is it possible to accelerate and not speed up or slow down?
Answer: No,
explanation: When the object is neither speeding up or slowing down, we can say that its speed is constant.
Hope this helps
Plz mark brainlesit
What are the benefits of living in a country with a growing population rate?
Answer:
The benefits are the place you live in will have more oppurtunities for new jobs as well as growth in residential areas.
Explanation:
when the temperature of matter decrease , the particles do what
Answer:
When the temperature decreases the particals start to slow down.
PLEASEEEEEE HELP
A jet makes a landing traveling due east with a speed of 120 m/s. If the jet comes to rest in 13.5 s , what’s the magnitude of its average acceleration?
Part B: What is the direction of its average acceleration? (North, south, east, west??)
Answer:
8.89 m/s² west
Explanation:
Assume east is +x. Given:
v₀ = 120 m/s
v = 0 m/s
t = 13.5 s
Find: a
v = at + v₀
0 m/s = a (13.5 s) + 120 m/s
a = -8.89 m/s²
a = 8.89 m/s² west
Starting from rest, a car accelerates at a rate of 7.8m/s^2 for 4.9 seconds. What is it’s velocity at the end of this time?
Answer:
7.8 m/'s = Change of accelerates / time taken =7.8/4.9=
Explanation:
please solve that answer
A falling object satisfies the initial value problem dv dt = 9.8 − v 5 , v(0) = 0 where v is the velocity in meters per second. (a) Find the time that must elapse for the object to reach 95% of its limiting velocity. (Round your answer to two decimal places.) s (b) How far does the object fall in the time found in part (a)? (Round your answer to two decimal places.) m Additional Materials
Answer:
a. t [tex]\simeq[/tex] 14.98 sec
b. x = 501.27 m
Explanation:
From the given information:
[tex]\dfrac{dv}{dt}=9.8-(\dfrac{v}{5 })[/tex] and [tex]v(0)=0[/tex]
[tex]\dfrac{dv}{dt}=\dfrac{49-v}{5 }[/tex]
[tex]\dfrac{dv}{49-v}=\dfrac{dt}{5 }[/tex]
Taking Integral of both sides
[tex]- ln(49-v) = \dfrac{t}{5} + C[/tex]
at t=0 we have v=0
This implies that
[tex]- ln(49-0) = \dfrac{0}{5} + C[/tex]
[tex]C= - ln(49)[/tex]
Thus:
[tex]\dfrac{t}{5} - In (49) = - In (49 -v) \\ \\ In(49) - \dfrac{t}{5} = In (49-v)[/tex]
[tex]49-v = e^{(-\frac{t}{5} +ln(49))}\\ \\ v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]
The limiting velocity when the time is infinite is :
95% of 49 = 46.55
∴
[tex]0.05= e^{(-\dfrac{t}{5})}[/tex]
[tex]\dfrac{t}{5}= In(\dfrac{1}{0.05})[/tex]
[tex]\dfrac{t}{5}=2.9957[/tex]
t = 5 × 2.9957
t [tex]\simeq[/tex] 14.98 sec
b.) [tex]v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]
[tex]v = \dfrac{dx}{dt}=49 - 49e^{(-\dfrac{t}{5})}[/tex]
[tex]dx=(49 - 49e^{(-\frac{t}{5})}) \ dt[/tex]
Taking integral of both sides.
[tex]x = 49t + 245 e^{(\frac{-t}{5})} +C[/tex]
at time t = 0 , distance x traveled = 0
∴
C= - 245
Therefore
[tex]x = 49t + 245 e^{(\frac{-t}{5})} -245[/tex]
replacing the value of t = 14.98
[tex]x = 49(14.98) + 245 e^{(\frac{-14.98}{5})} -245[/tex]
x = 501.27 m
The slope of the x-t curve at any point represents:_______a. Displacementb. Velocityc. Accerlationd. None of these
Answer:
Velocity
Explanation:
x-t curve means position vs time graph in which position (x) is in x-axis and time is in y-axis.
Slope of a graph = [tex]\dfrac{\Delta y}{\Delta t}[/tex].
We know that, velocity = displacement/time
or we can say that slope = displacement/time = velocity
Hence, the correct option is (b) "velocity".
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.A. Which laser has its first maximum closer to the central maximum?B. What is the distance delta ymax-min between the first maxima (on the same side of the central maximum) of the two patterns?C. What is the distance Delta ymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Answer:
Now
Using
ym = m x λ x L/d
therefore for each of first Max we have m =1
And laser 1 had
y = (d/20) x (6.0m)/d
y = 6.0m /20 = 0.3
Then laser 2 will be
y = 6.0m/15 = 0.4
y ( laser1 ) < y ( laser2 )
So the first maxima of laser 1 will be closer to the Central maxima
So
(0.4m -0 .3 m) = 0.1m
( C)
Now for laser 1 we say
y= 26.0 m / 20 = 0.6 m
Laser 2
We have
ym=(m+1/2) x λ x L/d
So
Because there is no central minimum the first minimum is at m = 0
We can way 3rd minimum is at m = 2
So
y = (2.5) x 6.0 / 15 = 1m
So
Δy= 1m - 0.6m = 0.4 m
A tennis ball is hit into the air and moves along an arc. (a) Neglecting air resistance, where along the arc is the speed of the ball a minimum? (Select all that apply.) at the initial position of motion at the highest point in the trajectory at the final position of motion
Answer:
at the highest point in the trajectory
Explanation:
When the tennis ball is hit, it moves in the air along a curve or an arc. This path is the parabola curve. Such a motion in the two dimension is known as projectile motion. It is constant accelerated motion in the downward direction.
The velocity of the ball is minimum at the highest point of the motion. When we hit the ball, the ball moves up to certain eight and then it gradually fall back to the earth surface along a curve.
The horizontal velocity of the ball is always the same along the curve. Only the vertical velocity varies. As the ball reaches the top of the curve or the maximum height, its vertical velocity becomes zero.
So, speed of the tennis ball is minimum at the highest point of the path.
A ball is thrown straight upward with a speed of 36 m/s. How long does it take to return to its starting point, assuming negligible air resistance?
Answer:
The time taken for the ball to return to the starting point is is 7.4 s
Explanation:
Given;
initial velocity of the ball, u = 36 m/s
the final vellocity at maximum height, v = 0
let time taken for the ball to reach maxmimum height = t
Time taken for the ball to return to the starting point is known as time of flight, calculated as;
[tex]t = \frac{v-u}{-g} \\\\T = 2t\\\\T = \frac{2(0-u)}{-g}\\\\T = \frac{-2u}{-g}\\\\ T = \frac{2u}{g}[/tex]
T = (2 x 36) / 9.8
T = 7.4 s
Therefore, the time taken for the ball to return to the starting point is is 7.4 s
If a moving clock is ticking half as fast as normal what speed is the clock traveling?
Answer:
Speed of moving clock is
[tex]V2[/tex] =πr/(86400) m/s
Here,
m and s are SI units of distance and time respectively.
Explanation:
If
radius of circular clock= r
than,
total distance covered on clock=S=2πr m
here, m=SI unit of distance
and required time for covering the total distance=t= 86400s
speed of normal clock=[tex]V1[/tex]=S/t
[tex]V1[/tex]=2πr/86400 m/s
As, moving moving clock is ticking half as fast as normal clock so,
speed of moving clock=[tex]V2[/tex]=[tex]V1[/tex]/2
[tex]V2[/tex]=2πr/(86400)*2 m/s
[tex]V2[/tex] =πr/(86400) m/s
HELLO CAN SOMEONE HELP ME PLS
A car is moving at 35 mph and comes to a stop in 5 seconds.
Find the acceleration of the car.
Answer:
I do believe it's 7
The morning after a massive snowstorm, Michaela gets into her car to drive to work. The storm caused her windows to freeze, so she first needs to defrost the car. While the engine is running, she checks the thermometer. It shows the air inside of her car has a temperature of 0 °C. Does this mean the air inside of her car has no kinetic energy? Explain your answer.
Answer:
Hope it helps
A Brainliest please
How does gamma radiation differ from alpha or beta particle radiation?
1) it does not consist of matter
2) it only consists of space
3) it does not consist of energy
4) it only consists of matter
Answer:
1.
Explanation:gamma rays are the most powerful in the electromagnetic spectrum and they are a result of a radioactive atom.they aren't made of matter but just energy as a wave.
4. What is the instantaneous acceleration at t= 10 s?
Answer:
I am fairly certain the answer is 2m/s^2
Explanation:
which of the following has the greatest inertia ping pong ball, golf ball, softball, and a bowling ball
Answer:
bowling ball
Explanation:
A bowling ball has more mass than the others, thus having more inertia.
In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min , 43.6 s43.6 s . Suppose "Pre" was at the 7.85 km7.85 km mark at a time of 25.0 min25.0 min . If he accelerated for 60 s60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s60 s interval. Assume his instantaneous speed at the 7.85 km7.85 km mark was the same as his overall average speed up to that time.
Answer:
a = 0.161 [tex]$m/s^2$[/tex]
Explanation:
Given :
[tex]$ d_{total}[/tex] = 10 km = 10000 m
[tex]$t_{total} $[/tex] = 27 min 43.6 s
= 1663.6 s
[tex]$d_1$[/tex] = 7.85 km = 7850 m
[tex]$t_1$[/tex] = 25 min = 1500 s
[tex]$t_2$[/tex] = 60 s
Now the initial speed for the distance of 7.85 km is
[tex]$ v_1 = \frac{d_1}{t_1} = \frac{7850}{1500}$[/tex] = 5.23 m/s
The velocity after 60 s after the distance of 7.85 kn is
[tex]$v_2 = v_1 + at_2$[/tex]
= 5.23 + a(60)
The distance traveled for 60 s after the distance of 7.85 km is
[tex]$d_2 = v_1t_2+\frac{1}{2}at_2^2$[/tex]
[tex]$d_2 = (5.23)(60)+\frac{1}{2}a(60)^2$[/tex]
= 313.8 + a(1800)
The time taken for the last journey where the speed is again uniform is
[tex]$t_3 = t_{total}-t_1-t_2 $[/tex]
= 1663.6 - 1500 - 60
= 103.6 s
Therefore, the distance traveled for the time [tex]$t_3$[/tex] is
[tex]$ d_3 = v_2 t_3$[/tex]
= (5.23+60a)(103.6)
= 541.8 + 6216 a
The total distance traveled,
[tex]$ d_{total}= d_1 + d_2 + d_3$[/tex]
Now substituting the values in the above equation for the acceleration a is
10000 = 7850 + (313.6 + 1800a) + (541.8 + 6216a)
10000 = 8706.5 + 8016a
1294.4 = 8016a
a = 0.161 [tex]$m/s^2$[/tex]
Which is not standing in the way of astronomers getting a good view of distant stars? A. some stars are too far away for our telescope to see B. lights get distorted by gas and dust as it goes through the atmosphere C. Redshift makes stars difficult to see D. there is alot of light pollution on earth
Answer: D. There is a lot of light pollution on earth
Explanation: The light pollution on Earth has nothing to do with the stars in the sky
Answer: C. Redshift makes stars difficult to see.
Explanation:
I did the test
You are walking at 3.75 km per hour across a frozen lake in the snow. You do not realize that with each step you turn 0.350 degrees to your right. If your step length is 74.0 cm what is the diameter, in meters, of the circle that you are inadvertently tracing out?
Answer:
242.27929622673 meters
